${K_{sp}}$ of $Pb$ $I_{2}$ is $ 1.4 \times {10^{ – 8}}$. The molecular mass of $Pb$ $I_{2}$ is $461$ $g$ $mol^{-1}$ Then molecular mass of $Pb{\left( {N{O_3}} \right)_2}$ is $331.9$ $mol^{-1}$ So, $(a)$ In $500$ $mL$ water $(b)$ $500$ $mL$ $0.10$ $M$ $KI$ $(c)$ What is the weight of $PbI_{2}$ when soluble in $1.33$ $g$ $Pb{\left( {N{O_3}} \right)_2}$ containing $500$ $mL$ solution ?

The molar solubility of $\mathrm{Zn}(\mathrm{OH})_{2}$ in $0.1 \,\mathrm{M} \,\mathrm{NaOH}$ solution is $\mathrm{x} \times 10^{-18} \,\mathrm{M}$. The value of $\mathrm{x}$ is …… . (Nearest integer)

(Given : The solubility product of $\mathrm{Zn}(\mathrm{OH})_{2}$ is $\left.2 \times 10^{-20}\right)$

If this is passed in solution of $0.1$ $M$ $Zn^{2+}$ and $0.01$ $M$ $Cu^{+}$ and concentration of ${S^{2 – }}$ made $8.1 \times {10^{ – 31}}$ $M$. Precipitation of $ZnS$ and $CuS$ will take place ? ${K_{sp}}$ of $\left( {ZnS} \right) = 3.0 \times {10^{ – 23}}$ and ${K_{sp}}$ of $\left( {CuS} \right) = 8.0 \times {10^{ – 34}}$

If the solubility product of $AgBr{O_3}$ and $A{g_2}S{O_4}$ are $5.5 \times {10^{ – 5}}$ and $2 \times {10^{ – 5}}$ respectively, the relationship between the solubilities of these can be correctly represented as

There are equal volume of $0.02$ $M$ $CaC{l_2}$ and $0.00004$ $M$ $N{a_2}S{O_4}$ solution are mixed will a precipitation of $CaS{O_4}$ ? ${K_{sp}} = 2.4 \times {10^{ – 5}}$